
1.1
Denote the set of all pentactions of \(A\) , \(A\in \mathbf {rGr}^{\bullet }\) , by \(\mathsf {Pentact}(A)\) .
Note that it follows from conditions \(2.\)  and \(2^{\circ }.\)  that \({0}^{b}={}^{b}{0}=0\) , and it follows from 4 that \({b}^{0}=0\) , for any \({(\text{-})}^{b},{\vphantom{(\text{-})}}^{b}{(\text{-})}\)  and \({b}^{(\text{-})}\in \mathbb {b}\) , \(\mathbb {b}\in \mathsf {Pentact}(A)\) .
Consider the set of functions
\(\mathbb {0}=\left(0\cdot \text{- },\text{-}\cdot 0,{(\text{-})}^{0},{\vphantom{(\text{-})}}^{0}{(\text{-})},{0}^{(\text{-})}\right),\) 
where \(0\cdot a = a\) , \(a\cdot 0=a\) , \({a}^{0}={}^{0}{a}=a\) , \({0}^{a}=0\) . It is easy to see that \(\mathbb {0}\)  is a pentaction; conditions 8 and \(8^{\circ }\)  hold for \(\mathbb {0}\) , since by definition \({a}^{0}={}^{0}{a}=a\)  for any \(a\in A\) .
For any \(\mathbb {b},\mathbb {b}^{\prime }\in \mathsf {Pentact}(A)\)  define a new set of functions, denoted by
\(\mathbb {b}+\mathbb {b}^{\prime }=\left(\left(b+b^{\prime }\right)\cdot \text{-}~,\text{-}\cdot \left(b+b^{\prime }\right),{(\text{-})}^{\left(b+b^{\prime }\right)},{\vphantom{(\text{-})}}^{\left(b+b^{\prime }\right)}{(\text{-})},{\left(b+b^{\prime }\right)}^{(\text{-})}\right),\) 
where by definition for any \(a\in A\)
\(\begin{alignedat}{2}\left(b+b^{\prime }\right)\cdot a &= b\cdot (b^{\prime }\cdot a) \\a\cdot \left(b+b^{\prime }\right) &= (a\cdot b)\cdot b^{\prime }\\{a}^{\left(b+b^{\prime }\right)} &= {\left({a}^{b}\right)}^{b^{\prime }}\\{\vphantom{a}}^{\left(b+b^{\prime }\right)}{a} &= {\vphantom{\left({\vphantom{a}}^{b^{\prime }}{a}\right)}}^{b}{\left({\vphantom{a}}^{b^{\prime }}{a}\right)}\\{\left(b+b^{\prime }\right)}^{a} &= {b}^{a} + b\cdot \left({b^{\prime }}^{a}\right).\end{alignedat}\) 
For any \(\mathbb {b},\mathbb {b}^{\prime }\in \mathsf {Pentact}(A)\)  consider a new set of functions
\({\mathbb {b}}^{\mathbb {b}^{\prime }}=\left(\left({b}^{b^{\prime }}\right)\cdot \text{- },\text{-}\cdot \left({b}^{b^{\prime }}\right),{(\text{-})}^{\left({b}^{b^{\prime }}\right)},{\vphantom{(\text{-})}}^{\left({b}^{b^{\prime }}\right)}{(\text{-})},{\left({b}^{b^{\prime }}\right)}^{(\text{-})}\right),\) 
defined by
\(\begin{alignedat}{2}\left({b}^{b^{\prime }}\right)\cdot a &= a \\a\cdot \left({b}^{b^{\prime }}\right) &= a\\{a}^{\left({b}^{b^{\prime }}\right)} &= {a}^{b}\\{\vphantom{a}}^{\left({b}^{b^{\prime }}\right)}{a} &= {\vphantom{a}}^{b}{a}\\{\left({b}^{b^{\prime }}\right)}^{a} &= {\left({b}^{\left(b^{\prime }\cdot a\right)}\right)}^{b^{\prime }},\end{alignedat}\) 
for any \(a\in A\) .
Definition 3.2 An object \(A\)  of the category \(\mathbf {rGr}^{\bullet }\)  will be called perfect if \({A}^{A}=A\) .

Here under \({A}^{A}\)  is meant the subobject of \(A\)  in \(\mathbf {rGr}^{\bullet }\)  generated by the elements of the type \({a}^{a^{\prime }}\) , \(a,a^{\prime }\in A\) .
Example 3.3 
Any abelian group with trivial action is perfect in \(\mathbf {rGr}^{\bullet }\) .

Example 3.4 For any \(A\in \mathbf {rGr}^{\bullet }\) , the object \({A}^{A}\)  is a perfect object. We have to show only that \({A}^{A}\subseteq {\left({A}^{A}\right)}^{\left({A}^{A}\right)}\) . Any element of \({A}^{A}\)  of the type \({{a}_{1}}^{{a}_{2}}\) , \({a}_{1},{a}_{2}\in A\) , can be represented as \({\left({{a}_{1}}^{{a}_{2}-y}\right)}^{\left({y}^{z}\right)}\)  for any \(y,z\in A\) , which means that \({{a}_{1}}^{{a}_{2}}\in {\left({A}^{A}\right)}^{\left({A}^{A}\right)}\) . The general case follows directly from this result, which proves that \({A}^{A}={\left({A}^{A}\right)}^{\left({A}^{A}\right)}\) .

This notion is an analogue of the well-known notions of perfect objects in the categories of associative or Leibniz algebras, which are algebras with representable actions, i.e. which have actors in the corresponding categories (see e.g. [1]}, [2]}).
Lemma 3.5 
If \(A\)  is a perfect object of \(\mathbf {rGr}^{\bullet }\) , then \(b\cdot a=a\)  for any \(b\cdot \text{-}\in \mathbb {b}\) , \(\mathbb {b}\in \mathsf {Pentact}(A)\) .

The result follows from the condition 6 of Definition REF .
Corollary 3.6 
\({\left({b}^{b^{\prime }}\right)}^{a}={\left({b}^{a}\right)}^{b^{\prime }}\) , for any \(a\in A\) , \({\left({b}^{b^{\prime }}\right)}^{(\text{-})}\in {\mathbb {b}}^{\mathbb {b}^{\prime }}\) , \(\mathbb {b},\mathbb {b}^{\prime }\in \mathsf {Pentact}(A)\) , \({b}^{(\text{-})}\in \mathbb {b}\) , \({(\text{-})}^{b^{\prime }}\in \mathbb {b}^{\prime }\) .

Here we apply the definition of \({\mathbb {b}}^{\mathbb {b}^{\prime }}\) , where
\({\left({b}^{b^{\prime }}\right)}^{a}={\left({b}^{b^{\prime }\cdot a}\right)}^{b^{\prime }}\) 
and by Lemma REF  we have \(b^{\prime }\cdot a =a\) .
Corollary 3.7 \({\left({a}^{a^{\prime }}\right)}^{b}={\left({a}^{b}\right)}^{a^{\prime }}\) , where \({(\text{-})}^{b}\in \mathbb {b}\) , \(\mathbb {b}\in \mathsf {Pentact}(A)\) .

It follows from the condition 5 of Definition REF  and Lemma REF .
Definition 3.8 A stabilizer of the object \(A\in \mathbf {rGr}^{\bullet }\) , denoted by \(\mathsf {Stab}(A)\) , is a set of elements \(\overline{a}\in A\)  with the property that \({a}^{\overline{a}}=a\) , for any \(a\in A\) , i.e.
\(\mathsf {Stab}(A)=\left\lbrace \overline{a}\in A \;|\; {a}^{\overline{a}}=a, \text{ for any } a\in A \right\rbrace .\) 

Note that a notion of stabilizer of an element of a set is known for the case group with action on a set. We will see that annihilation of certain type stabilizers, defined below, will play analogous role in the category of reduced groups with action, as annihilators play in the case of associative and Leibniz algebras for the existence of actors in these categories; here we mean the well-known condition \(\mathsf {Ann}(A)=0\)  in the corresponding categories (see e.g. [1]}, [2]}).
Lemma 3.9 Let \(A\in \mathbf {rGr}^{\bullet }\) . If \(\mathsf {Stab}(A)=0\) , then \(A=0\) .

For any \(a,a_1\)  and \(a_2\)  from \(A\)  we have
\({a}^{\left({a_{1}}^{a_2}-a_1\right)}={\left({a}^{a_1}\right)}^{-a_1}=a;\) 
therefore \({a_1}^{a_2}-a_1=0\) , since \(\mathsf {Stab}(A)=0\) . From this it follows that \({a_1}^{a_2}=a_1\) , which means that the action of \(A\)  on itself is trivial; it implies that \(A=\mathsf {Stab}(A)=0\) .
Definition 3.10 A weak stabilizer of an object \(A\in \mathbf {rGr}^{\bullet }\)  is defined as follows;
\(\mathsf {wStab}(A)=\left\lbrace {b}^{\left({b^{\prime }}^{a}\right)}, {b}^{\left({a}^{b^{\prime }}\right)}-{b}^{a}, {a}^{b+b^{\prime }}-{a}^{b^{\prime }+b}\right\rbrace ,\) 
where \(a\in A\) , \({b}^{(\text{-})}\in \mathbb {b}\) , \({b^{\prime }}^{(\text{-})},{(\text{-})}^{b^{\prime }}\in \mathbb {b}^{\prime }\) , \({(\text{-})}^{b+b^{\prime }}\in \mathbb {b}+\mathbb {b}^{\prime }\) , \({(\text{-})}^{b^{\prime }+b}\in \mathbb {b}^{\prime }+\mathbb {b}\) , \(\mathbb {b},\mathbb {b}^{\prime }\in \mathsf {Pentact}(A)\) .

Lemma 3.11 \(\mathsf {wStab}(A)\)  is a subset of \(\mathsf {Stab}(A)\) , for any \(A\in \mathbf {rGr}^{\bullet }\) .

For any \(\overline{a}\in A\) , we have \({\overline{a}}^{\left({b}^{\left({b^{\prime }}^{a}\right)}\right)}=\overline{a}\)  by Definition REF , condition 10. Applying the same condition we have
\({\overline{a}}^{\left({b}^{\left({a}^{b^{\prime }}\right)}-{b}^{a}\right)} = {\left({\overline{a}}^{\left({b}^{\left({a}^{{b}^{\prime }}\right)}\right)}\right)}^{\left(-{b}^{a}\right)} = {\vphantom{\left(A^a\right)} \overline{a}}^{\left(-{b}^{a}\right)} = \overline{a}.\) 
Here we applied that if \({\vphantom{\left(A^a\right)} \overline{a}}^{{a}_{1}}=\overline{a}\) , then it follows that \({\vphantom{\left(A^a\right)} \overline{a}}^{\left(-{a}_{1}\right)}=\overline{a}\) .
We have the following equalities
\({\vphantom{\left(A^a\right)}\overline{a}}^{\left({a}^{\left(b+b^{\prime }\right)}-{a}^{\left(b^{\prime }+b\right)}\right)} = {\left({\vphantom{\left(A^a\right)} \overline{a}}^{a}\right)}^{-a} = {\vphantom{\left(A^a\right)} \overline{a}}^{0} = a,\) 
where we applied condition 9 of Definition REF  and the fact that \(-\left({a}^{{b}_{_1}}\right)={(-a)}^{{b}_{1}}\) , since
\(0={0}^{{b}_{1}}={(-a+a)}^{{b}_{1}}={(-a)}^{{b}_{1}}+{a}^{{b}_{1}},\) 
for any \({(\text{-})}^{{b}_{1}}\in {\mathbb {b}}_{1}\) , \({\mathbb {b}}_{1}\in \mathsf {Pentact}(A)\) .
Note that \(\mathsf {Stab}(A)\)  is a subobject of \(A\) , while \(\mathsf {wStab}(A)\)  is a subset of \(A\)  in general.
Proposition 3.12 
Let \(A\in \mathbf {rGr}^{\bullet }\)  and \(\mathbb {b},\mathbb {b}^{\prime }\in \mathsf {Pentact}(A)\) . If \(A\)  is perfect in \(\mathbf {rGr}^{\bullet }\) , then \(\mathbb {b}+\mathbb {b}^{\prime }\in \mathsf {Pentact}(A)\) .

We shall show that \(\mathbb {b}+\mathbb {b}^{\prime }\)  defined by (REF ) satisfies conditions 1–12 of Definition REF .

Checking of condition 1: for the left side we have
\(\left(b+b^{\prime }\right)\cdot (a+a^{\prime })&=b\cdot (b^{\prime }\cdot (a+a^{\prime }))\\&=b\cdot (b^{\prime }\cdot a+b^{\prime }\cdot a^{\prime }) \\&=b\cdot (b^{\prime }\cdot a)+ b\cdot (b^{\prime }\cdot a^{\prime }) \\&= \left(b+b^{\prime }\right)\cdot a + \left(b+b^{\prime }\right)\cdot a^{\prime },\) 

which proves the condition.

Condition 2: for the left side we have
\({(a+a^{\prime })}^{\left(b+b^{\prime }\right)}&= {\left({(a+a^{\prime })}^{b}\right)}^{b^{\prime }} \\&= {\left({a}^{b}\right)}^{b^{\prime }}+{\left({a^{\prime }}^{b}\right)}^{b^{\prime }} \\&= {a}^{\left(b+b^{\prime }\right)}+{a^{\prime }}^{\left(b+b^{\prime }\right)},\) 

which ends the proof.

For \(a\ne 0\)  we have
\({\left(\left(b+b^{\prime }\right)\cdot a\right)}^{a^{\prime }} = {\left(b\cdot (b^{\prime }\cdot a)\right)}^{a^{\prime }} = {\left(b^{\prime }\cdot a\right)}^{a^{\prime }} = {a}^{a^{\prime }},\) 
here we applied that condition 3 holds for \(b\)  and \(b^{\prime }\) , and therefore proved that this condition for any \(A\) , not taking into account that
\(A\)  is perfect.

We have to show
\({\left(b+b^{\prime }\right)}^{(a+a^{\prime })}={\left({\left(b+b^{\prime }\right)}^{a}\right)}^{a^{\prime }}+{\left(b+b^{\prime }\right)}^{a^{\prime }}.\) 
For the left side we have
\({\left(b+b^{\prime }\right)}^{(a+a^{\prime })} &= {b}^{a+a^{\prime }}+b\cdot {b^{\prime }}^{a+a^{\prime }} \\&= {\left({b}^{a}\right)}^{a^{\prime }}+{b}^{a^{\prime }}+b\cdot \left({\left({b^{\prime }}^{a}\right)}^{a^{\prime }}+{b^{\prime }}^{a^{\prime }}\right) \\&= {\left({b}^{a}\right)}^{a^{\prime }}+{b}^{a^{\prime }}+b\cdot {\left({b^{\prime }}^{a}\right)}^{a^{\prime }}+b\cdot \left({b^{\prime }}^{a^{\prime }}\right).\) 

For the right side we have
\({\left({\left(b+b^{\prime }\right)}^{a}\right)}^{a^{\prime }}+{\left(b+b^{\prime }\right)}^{a^{\prime }} &= {\left({b}^{a}+b\cdot {b^{\prime }}^{a}\right)}^{a^{\prime }}+{b}^{a^{\prime }}+b\cdot \left({b^{\prime }}^{a^{\prime }}\right) \\&= {\left({b}^{a}\right)}^{a^{\prime }}+{\left(b\cdot {b^{\prime }}^{a}\right)}^{a^{\prime }}+{b}^{a^{\prime }}+b\cdot \left({b^{\prime }}^{a^{\prime }}\right).\) 

Since \(A\)  is perfect \({\left(b\cdot {b^{\prime }}^{a}\right)}^{a^{\prime }}+{b}^{a^{\prime }}={b}^{a^{\prime }}+{\left(b\cdot {b^{\prime }}^{a}\right)}^{a^{\prime }}\)  which ends the proof of equality 4.

In order to prove this condition we need to show that the following equality is satisfied.
\({\left({a}^{\left(b+b^{\prime }\right)\cdot a^{\prime }}\right)}^{\left(b+b^{\prime }\right)}={\left({a}^{\left(b+b^{\prime }\right)}\right)}^{a^{\prime }}.\) 
We compute first the left side, which is equal to the following
\({\left({a}^{b\cdot (b^{\prime }\cdot a^{\prime })}\right)}^{\left(b+b^{\prime }\right)} = {\left({\left({a}^{b\cdot (b^{\prime }\cdot a^{\prime })}\right)}^{b}\right)}^{b^{\prime }} = {\left({\left({a}^{b}\right)}^{b^{\prime }\cdot a^{\prime }}\right)}^{b^{\prime }}.\) 
Here we applied condition 5 for \({(\text{-})}^{b}\)  and \({(\text{-})}^{b^{\prime }}\) . Note that one could apply also the fact that \(A\)  is perfect, and replace \(b^{\prime }\cdot a^{\prime }\)  by \(a^{\prime }\)  by Lemma REF , but we will show that this condition holds for \(\mathbb {b}+\mathbb {b}^{\prime }\)  for any \(A\in \mathbf {rGr}^{\bullet }\) . The right side is equal to
\({\left({\left({a}^{b}\right)}^{b^{\prime }}\right)}^{a^{\prime }} = {\left({\left({a}^{b}\right)}^{b^{\prime }\cdot a^{\prime }}\right)}^{b^{\prime }},\) 
here we applied condition 5 for \({(\text{-})}^{b^{\prime }}\) , which proves the desired equality.

We have to show that \(\left(b+b^{\prime }\right)\cdot {a}^{a^{\prime }}={a}^{a^{\prime }}\)  for \(a^{\prime }\ne 0\) . The left side is equal to \(b\cdot \left(b^{\prime }\cdot {a}^{a^{\prime }}\right)=b\cdot {a}^{a^{\prime }}={a}^{a^{\prime }}\) , here we applied condition 6 for \(b^{\prime }\cdot \text{-}\)  and \(b\cdot \text{-}\) .

Now we will prove the equality
\({\left(b+b^{\prime }\right)}^{\left({a}^{a^{\prime }}\right)}={\left(b+b^{\prime }\right)}^{a}.\) 
We have
\({\left(b+b^{\prime }\right)}^{\left({a}^{a^{\prime }}\right)} &= {b}^{\left({a}^{a^{\prime }}\right)}+b\cdot {b^{\prime }}^{\left({a}^{a^{\prime }}\right)} \\&= {b}^{a}+b\cdot {(b^{\prime })}^{a} \\&={\left(b+b^{\prime }\right)}^{a},\) 

since \({b}^{(\text{-})}\)  and \({b^{\prime }}^{(\text{-})}\)  satisfy condition 7, which proves the equality.

The equality
\({a}^{\left(b+b^{\prime }\right)}+a^{\prime }=a^{\prime }+{a}^{\left(b+b^{\prime }\right)}.\) 
always holds since \(A\)  is perfect.

We need to show that
\({a}^{\left({a^{\prime }}^{\left(b+b^{\prime }\right)}\right)}={a}^{a^{\prime }}.\) 
The left side is equal to
\({a}^{\left(\left({a^{\prime }}^{b}\right)^{b^{\prime }}\right)} = {a}^{\left({a^{\prime }}^{b}\right)} = {a}^{a^{\prime }}.\) 
Here we applied condition 9 for \({(\text{-})}^{b}\)  and \({(\text{-})}^{b^{\prime }}\) .

Now we check the condition
\({a}^{\left({\left(b+b^{\prime }\right)}^{a^{\prime }}\right)}=a.\) 
We compute the left side, which is equal to
\({a}^{\left({b}^{a^{\prime }}+b\cdot {b^{\prime }}^{a^{\prime }}\right)}={\left({a}^{\left({b}^{a^{\prime }}\right)}\right)}^{\left(b\cdot {b^{\prime }}^{a^{\prime }}\right)},\) 
here we will apply that \(A\)  is perfect and the fact that \({a}^{\left({b}^{a^{\prime }}\right)}=a\)  by condition 10 for \(\mathbb {b}\) , which gives that the obtained expression is equal to \({a}^{\left({b^{\prime }}^{a^{\prime }}\right)}=a\) , since \(\mathbb {b}^{\prime }\)  satisfies condition 10.

We have to show
\((\left(b+b^{\prime }\right)\cdot a)\cdot \left(b+b^{\prime }\right)=\left(b+b^{\prime }\right)\cdot (a\cdot \left(b+b^{\prime }\right))=a.\) 
We apply that this condition holds for \(\mathbb {b}\)  and \(\mathbb {b}^{\prime }\)  and obtain that the left side is equal to
\(\left(\left(b\cdot \left(b^{\prime }\cdot a\right)\right)\cdot b\right)\cdot b^{\prime } = \left(b^{\prime }\cdot a\right)\cdot b^{\prime }=a.\) 
Note that this equality is proved for any \(A\) . If we take into account that \(A\)  is perfect, then it is trivial that the equality holds, since \(\mathbb {b}\)  and \(\mathbb {b}^{\prime }\)  satisfy conditions 6 and 6\(^{circ}\)  of Definition REF .

Last of all we prove the equality
\({\vphantom{\left({}^{\left(b+b^{\prime }\right)}{a}\right)}}^{\left(b+b^{\prime }\right)}{\left({a}^{\left(b+b^{\prime }\right)}\right)}={\left({}^{\left(b+b^{\prime }\right)}{a}\right)}^{\left(b+b^{\prime }\right)}=a.\) 
The left side is equal to
\({\vphantom{\left({\vphantom{\left({\left({a}^{b}\right)}^{b^{\prime }}\right)}}^{b^{\prime }}{\left({\left({a}^{b}\right)}^{b^{\prime }}\right)}\right)}}^{b}{\left({\vphantom{\left({\left({a}^{b}\right)}^{b^{\prime }}\right)}}^{b^{\prime }}{\left({\left({a}^{b}\right)}^{b^{\prime }}\right)}\right)}={\vphantom{\left({a}^{b}\right)}}^{b}{\left({a}^{b}\right)}=a,\) 
where we apply condition 12 for \(\mathbb {b}\)  and \(\mathbb {b}^{\prime }\) .

The equalities \(1^{\circ },2^{\circ },3^{\circ },5^{\circ },6^{\circ },8^{\circ }\)  and \(9^{\circ }\)  are proved in the similar ways as the equalities \(1,2,3,5,6,8\)  and 9, respectively.
Proposition 3.13 For any \(A\in \mathbf {rGr}^{\bullet }\)  and \(\mathbb {b},\mathbb {b}^{\prime }\in \mathsf {Pentact}(A)\) , if \(A\)  is perfect, then \({\mathbb {b}}^{\mathbb {b}^{\prime }}\in \mathsf {Pentact}(A)\) .

We will check the conditions of Definition REF  like we did it in the proof of Proposition REF .

We have \({b}^{b^{\prime }}\cdot (a+a^{\prime })={b}^{b^{\prime }}\cdot a+{b}^{b^{\prime }}\cdot a^{\prime }\) , since by the definition of \({\mathbb {b}}^{\mathbb {b}^{\prime }}\) , \({b}^{b^{\prime }}\cdot a=a\)  for any \(a\in A\) .

We have to show
\({(a+a^{\prime })}^{\left({b}^{b^{\prime }}\right)}={a}^{\left({b}^{b^{\prime }}\right)}+{a^{\prime }}^{\left({b}^{b^{\prime }}\right)}.\) 
The left side is equal to \({(a+a^{\prime })}^{b}\)  by the definition of \({\mathbb {b}}^{\mathbb {b}^{\prime }}\)  and we have \({(a+a^{\prime })}^{b}={a}^{b}+{a^{\prime }}^{b}\) , since \(\mathbb {b}\in \mathsf {Pentact}(A)\) . The right side is equal to the same expression again by the definition of \({\mathbb {b}}^{\mathbb {b}^{\prime }}\) .

For any \(a^{\prime }\ne 0\)  we have to show
\({({b}^{b^{\prime }}\cdot a)}^{a^{\prime }}={a}^{a^{\prime }}.\) 
The left side is equal to \({a}^{a^{\prime }}\)  by the definition of \({\mathbb {b}}^{\mathbb {b}^{\prime }}\) .

By the definition of \({\mathbb {b}}^{\mathbb {b}^{\prime }}\)  the left side is equal to
\({\left({b}^{b^{\prime }}\right)}^{(a+a^{\prime })}&= {\left({b}^{b^{\prime }\cdot (a+a^{\prime })}\right)}^{b^{\prime }} \\&= {\left({b}^{b^{\prime }\cdot a+b^{\prime }\cdot a^{\prime }}\right)}^{b^{\prime }} \\&= {\left({\left({b}^{b^{\prime }\cdot a}\right)}^{b^{\prime }\cdot a^{\prime }}+{b}^{b^{\prime }\cdot a^{\prime }}\right)}^{b^{\prime }} \\&= {\left({\left({b}^{b^{\prime }\cdot a}\right)}^{b^{\prime }}\right)}^{a^{\prime }} + {\left({b}^{b^{\prime }\cdot a^{\prime }}\right)}^{b^{\prime }},\) 

here we applied condition 4 of Definiton REF . By the definition of \({\mathbb {b}}^{\mathbb {b}^{\prime }}\)  the right side is equal to \({\left({\left({b}^{b^{\prime }\cdot a}\right)}^{b^{\prime }}\right)}^{a^{\prime }} + {\left({b}^{b^{\prime }\cdot a^{\prime }}\right)}^{b^{\prime }}\) , which proves the equality.

In order to prove this condition we need to show that the following equality is satisfied.
\({\left({a}^{\left({b}^{b^{\prime }}\right)\cdot a^{\prime }}\right)}^{\left({b}^{b^{\prime }}\right)}={\left({a}^{\left({b}^{b^{\prime }}\right)}\right)}^{a^{\prime }}.\) 
The left side is equal to
\({\left({a}^{a^{\prime }}\right)}^{\left({b}^{b^{\prime }}\right)}={\left({a}^{a^{\prime }}\right)}^{b}.\) 
The right side is equal to
\({\left({a}^{b}\right)}^{a^{\prime }}={\left({a}^{b\cdot a^{\prime }}\right)}^{b}={\left({a}^{a^{\prime }}\right)}^{b};\) 
here we applied that \(A\)  is perfect and therefore \(b\cdot a^{\prime }=a^{\prime }\) , which proves the equality.

We have \({b}^{b^{\prime }}\cdot {a}^{a^{\prime }}={a}^{a^{\prime }}\)  for \(a^{\prime }\ne 0\) , by the definition of \({\mathbb {b}}^{\mathbb {b}^{\prime }}\) .

Now we will prove the equality
\({\left({b}^{b^{\prime }}\right)}^{\left({a}^{a^{\prime }}\right)}={\left({b}^{b^{\prime }}\right)}^{a}.\) 
The left side is equal to
\({\left({b}^{b^{\prime }\cdot \left({a}^{a^{\prime }}\right)}\right)}^{b^{\prime }}={\left({b}^{\left({a}^{a^{\prime }}\right)}\right)}^{b^{\prime }}={\left({b}^{a}\right)}^{b^{\prime }},\) 
here we applied conditions 6 and 7 of Definition REF . The right side is equal to
\({\left({b}^{b^{\prime }\cdot a}\right)}^{b^{\prime }}={\left({b}^{a}\right)}^{b^{\prime }},\) 
since \(A\)  is perfect and \(b^{\prime }\cdot a=a\) , which proves the equality.

We have to show that if \({a}^{\left({b}^{b^{\prime }}\right)}\ne a\)  for at least one \(a\in A\) , then
\({a}^{\left({b}^{b^{\prime }}\right)}+a^{\prime }=a^{\prime }+{a}^{\left({b}^{b^{\prime }}\right)}.\) 
The condition \({a}^{\left({b}^{b^{\prime }}\right)}\ne a\)  for at least one \(a\in A\)  is equivalent to the condition that \({a}^{b}\ne a\)  for at least one \(a\in A\) .
The left side is equal to
\({a}^{\left({b}^{b^{\prime }}\right)}+a^{\prime }={a}^{b}+a^{\prime },\) 
by the definition of \({\mathbb {b}}^{\mathbb {b}^{\prime }}\)  and the condition 8 of Definition REF  for \(\mathbb {b}\in \mathsf {Pentact}(A)\) . Note that the equality 8 is true in the case when \(A\)  is not perfect.

We have to show that
\({a}^{\left({a^{\prime }}^{\left({b}^{b^{\prime }}\right)}\right)}={a}^{a^{\prime }}.\) 
For the left side, by the definition of \({\mathbb {b}}^{\mathbb {b}^{\prime }}\) , we have
\({a}^{\left({a^{\prime }}^{\left({b}^{b^{\prime }}\right)}\right)}={a}^{\left({a^{\prime }}^{b}\right)}={a}^{a^{\prime }},\) 
since \(\mathbb {b}\)  is a pentaction.

Now we check the condition
\({a}^{\left({\left({b}^{b^{\prime }}\right)}^{a^{\prime }}\right)}=a.\) 
We have
\({a}^{\left({\left({b}^{b^{\prime }}\right)}^{a^{\prime }}\right)}={a}^{\left({\left({b}^{\left(b^{\prime }\cdot a^{\prime }\right)}\right)}^{b^{\prime }}\right)}={a}^{\left({b}^{\left(b^{\prime }\cdot a^{\prime }\right)}\right)}=a.\) 
Here we applied the definition of \({\mathbb {b}}^{\mathbb {b}^{\prime }}\) , conditions 9 and 10 for \(\mathbb {b}\)  and \(\mathbb {b}^{\prime }\) , respectively, of Definition REF .

Condition 11 for \({\mathbb {b}}^{\mathbb {b}^{\prime }}\)  follows directly from the definition of \({\mathbb {b}}^{\mathbb {b}^{\prime }}\) . Condition 12 follows as well from the definition of \({\mathbb {b}}^{\mathbb {b}^{\prime }}\)  and the fact that \(\mathbb {b}\)  is a pentaction with property 12.
The conditions \(1^{\circ },2^{\circ },3^{\circ },5^{\circ },6^{\circ },8^{\circ }\)  and \(9^{\circ }\)  are proved in the similar ways as their duals above.
Theorem 3.14 
If \(A\)  is a perfect object of \(\mathbf {rGr}^{\bullet }\)  with \(\mathsf {wStab}(A)=0\) , then \(\mathsf {Pentact}(A)\)  is an object of \(\mathbf {rGr}^{\bullet }\) .

First we have to show, that the addition defined in \(\mathsf {Pentact}(A)\)  is associative, the function \(\mathbb {0}\)  defined by us is the zero element in this set and there exists an opposite element for any \(\mathbb {b}\in \mathsf {Pentact}(A)\) . After this we will show that the action conditions are satisfied in \(\mathsf {Pentact}(A)\)  and that we have two additional conditions as for reduced groups with actions presented in Section .
It is easy to see that the addition is associative in \(\mathsf {Pentact}(A)\) , i.e. \(\left(\mathbb {b}+\mathbb {b}^{\prime }\right)+\mathbb {b}^{\prime \prime }=\mathbb {b}+\left(\mathbb {b}^{\prime }+\mathbb {b}^{\prime \prime }\right)\) , for any \(\mathbb {b},\mathbb {b}^{\prime },\mathbb {b}^{\prime \prime }\in \mathsf {Pentact}(A)\) . It is obvious that the zero element is the function \(\mathbb {0}\)  defined by us. For any \(\mathbb {b}\in \mathsf {Pentact}(A)\)  the opposite element is defined by
\(-\mathbb {b}=\left(\left(-b\right)\cdot \text{- } , \text{-} \cdot \left(-b\right) , {(\text{-})}^{\left(-b\right)},{\vphantom{(\text{-})}}^{\left(-b\right)}{(\text{-})},{\left(-b\right)}^{(\text{-})}\right),\) 
where
\(\left(-b\right)\cdot \text{-} &= \text{-} \cdot b \\\text{-} \cdot \left(-b\right) &= b\cdot \text{-} \\{(\text{-})}^{\left(-b\right)} &= {\vphantom{(\text{-})}}^{b}{(\text{-})} \\{\vphantom{(\text{-})}}^{\left(-b\right)}{(\text{-})} &= {(\text{-})}^{b} \\{\left(-b\right)}^{(\text{-})} &= -\left[\left({b}^{(\text{-})}\right)\cdot b\right].\) 

It is easy to check that \(\mathbb {b}+\left(-\mathbb {b}\right)=-\mathbb {b}+\mathbb {b}=\mathbb {0}\) .
Now we have to show that the addition defined by us in \(\mathsf {Pentact}(A)\)  satisfies action conditions in the category of groups.
Therefore we have to show

\({\left(\mathbb {b}+\mathbb {b}^{\prime }\right)}^{\mathbb {b}^{\prime \prime }}={\mathbb {b}}^{\mathbb {b}^{\prime \prime }}+{\mathbb {b}^{\prime }}^{\mathbb {b}^{\prime \prime }}\)

\({\mathbb {b}}^{\left(\mathbb {b}^{\prime }+\mathbb {b}^{\prime \prime }\right)}={\left({\mathbb {b}}^{\mathbb {b}^{\prime }}\right)}^{\mathbb {b}^{\prime \prime }}\)

\({\mathbb {b}}^{\mathbb {0}}=\mathbb {b}\)

for any \(\mathbb {b},\mathbb {b}^{\prime },\mathbb {b}^{\prime \prime }\in \mathsf {Pentact}(A)\) .

We have \({\left(b+b^{\prime }\right)}^{b^{\prime \prime }}\cdot a=a=\left({b}^{b^{\prime \prime }}+{b^{\prime }}^{b^{\prime \prime }}\right)\cdot a\)  for any \({\left(b+b^{\prime }\right)}^{b^{\prime \prime }}\cdot \text{-}\in {\left(\mathbb {b}+\mathbb {b}^{\prime }\right)}^{\mathbb {b}^{\prime \prime }}\) , any \(\left({b}^{b^{\prime \prime }}+{b^{\prime }}^{b^{\prime \prime }}\right)\cdot \text{-}\in {\mathbb {b}}^{\mathbb {b}^{\prime \prime }}+{\mathbb {b}^{\prime }}^{\mathbb {b}^{\prime \prime }}\)  and any \(a\in A\) .
We have to show
\({a}^{\left({\left(b+b^{\prime }\right)}^{b^{\prime \prime }}\right)}={a}^{\left({b}^{b^{\prime \prime }}+{b^{\prime }}^{b^{\prime \prime }}\right)}.\) 
The left side is equal to \({\left({a}^{b}\right)}^{b^{\prime }}={a}^{\left({b}^{b^{\prime \prime }}+{b^{\prime }}^{b^{\prime \prime }}\right)}\) , which proves the equality. Moreover we have to show
\({\left({\left(b+b^{\prime }\right)}^{b^{\prime \prime }}\right)}^{a}={\left({b}^{b^{\prime \prime }}+{b^{\prime }}^{b^{\prime \prime }}\right)}^{a}.\) 
The left side is equal to
\({\left({\left(b+b^{\prime }\right)}^{(b^{\prime \prime }\cdot a)}\right)}^{b^{\prime \prime }} &= {\left({b}^{a}+b\cdot {b^{\prime }}^{a}\right)}^{b^{\prime \prime }} \\&= {\left({b}^{a}\right)}^{b^{\prime \prime }}+{\left(b\cdot {b^{\prime }}^{a}\right)}^{b^{\prime \prime }} \\&= {\left({b}^{a}\right)}^{b^{\prime \prime }}+{\left({b^{\prime }}^{a}\right)}^{b^{\prime \prime }},\) 

here we applied Lemma REF  since \(A\)  is perfect.
The right side is equal to
\({\left({b}^{b^{\prime \prime }}\right)}^{a}+{b}^{b^{\prime \prime }}\cdot {\left({b^{\prime }}^{b^{\prime \prime }}\right)}^{a} &= {\left({b}^{b^{\prime \prime }}\right)}^{a}+{\left({b^{\prime }}^{b^{\prime \prime }}\right)}^{a} \\&= {\left({b}^{a}\right)}^{b^{\prime \prime }}+{\left({b^{\prime }}^{a}\right)}^{b^{\prime \prime }},\) 

here we applied Corollary REF  which proves the desired equality.
The equalities for the dual actions are proved similarly.

We have to show
\({b}^{\left(b^{\prime }+b^{\prime \prime }\right)}\cdot a = {\left({b}^{b^{\prime }}\right)}^{b^{\prime \prime }}\cdot a,\) 
for any \({b}^{\left(b^{\prime }+b^{\prime \prime }\right)}\cdot \text{-}\in {\mathbb {b}}^{\mathbb {b}^{\prime }+\mathbb {b}^{\prime \prime }}\)  and \({\left({b}^{b^{\prime }}\right)}^{b^{\prime \prime }}\cdot \text{-}\in {\left({\mathbb {b}}^{\mathbb {b}^{\prime }}\right)}^{\mathbb {b}^{\prime \prime }}\) . Both sides are equal to \(a\)  by the definition of action operation in \(\mathsf {Pentact}(A)\) . The equality
\({a}^{\left({b}^{\left(b^{\prime }+b^{\prime \prime }\right)}\right)}={a}^{\left({\left({b}^{b^{\prime }}\right)}^{b^{\prime \prime }}\right)}\) 
holds by the definition of action operation ib \(\mathsf {Pentact}(A)\) .
Now we have to show
\({\left({b}^{\left(b^{\prime }+b^{\prime \prime }\right)}\right)}^{a}={\left({\left({b}^{b^{\prime }}\right)}^{b^{\prime \prime }}\right)}^{a}.\) 
Both sides are equal to \({\left({\left({b}^{\left(b^{\prime }\cdot \left(b^{\prime \prime }\cdot a\right)\right)}\right)}^{b^{\prime }}\right)}^{b^{\prime \prime }}\)  without application of the condition that \(A\)  is perfect, which gives the desired equality.

We have to show
\({b}^{0}\cdot a = b\cdot a.\) 
The left side is equal to \(a\)  by the definition of action operation in \(\mathsf {Pentact}(A)\) , and the right side is equal to \(a\)  as well by Lemma REF  since \(A\)  is perfect.
\({a}^{\left({b}^{0}\right)}={a}^{b}\) 
by the definition of action \({\mathbb {b}}^{\mathbb {0}}\) .
We have to show that
\({\left({b}^{0}\right)}^{a}={b}^{a}.\) 
The left side is equal to
\({\left({b}^{0\cdot a}\right)}^{a}={\left({b}^{a}\right)}^{0}={b}^{a},\) 
which gives the equality. For the dual actions the corresponding equalities are proved similarly.

Now we have to show that in \(\mathsf {Pentact}(A)\)  we have

\({\mathbb {b}}^{\mathbb {b}^{\prime }}+\mathbb {b}^{\prime \prime }=\mathbb {b}^{\prime \prime }+{\mathbb {b}}^{\mathbb {b}^{\prime }}\)  for \(\mathbb {b}^{\prime }\ne \mathbb {0}\)

\({\mathbb {b}}^{\left({\mathbb {b}^{\prime }}^{\mathbb {b}^{\prime \prime }}\right)}={\mathbb {b}}^{\mathbb {b}^{\prime }}\)

for any \(\mathbb {b},\mathbb {b}^{\prime }\)  and \(\mathbb {b}^{\prime \prime }\in \mathsf {Pentact}(A)\) .

We have to show
\(\left({b}^{b^{\prime }}+b^{\prime \prime }\right)\cdot a = \left(b^{\prime \prime }+{b}^{b^{\prime }}\right)\cdot a,\) 
for any \(\left({b}^{b^{\prime }}+b^{\prime \prime }\right)\cdot \text{-}\in {\mathbb {b}}^{\mathbb {b}^{\prime }}+\mathbb {b}^{\prime \prime }\) , any \(\left(b^{\prime \prime }+{b}^{b^{\prime }}\right)\cdot \text{-}\in \mathbb {b}^{\prime \prime }+{\mathbb {b}}^{\mathbb {b}^{\prime }}\)  such that \(\mathbb {b}^{\prime }\ne \mathbb {0}\) , and any \(a\in A\) .
The both sides are equal to \(b^{\prime \prime }\cdot a\)  by the definition, and the equality holds for any \(A\) , without being perfect.
We have to show
\({a}^{\left({b}^{b^{\prime }}+b^{\prime \prime }\right)}={a}^{\left(b^{\prime \prime }+{b}^{b^{\prime }}\right)}.\) 
The left side is equal to
\({\left({a}^{\left({b}^{b^{\prime }}\right)}\right)}^{b^{\prime \prime }}={\left({a}^{b}\right)}^{b^{\prime \prime }}={a}^{b+b^{\prime \prime }}.\) 
And the right side is equal to \({a}^{b^{\prime \prime }+b}\) . Here we apply that \(\mathsf {wStab}(A)=0\) , which implies that these two elements are equal.
Moreover we have to show
\({\left({b}^{b^{\prime }}+b^{\prime \prime }\right)}^{a}={\left(b^{\prime \prime }+{b}^{b^{\prime }}\right)}^{a}.\) 
The left side is equal to
\({\left({b}^{b^{\prime }}\right)}^{a}+{b}^{b^{\prime }}\cdot \left({b^{\prime \prime }}^{a}\right)={\left({b}^{b^{\prime }}\right)}^{a}+{b^{\prime \prime }}^{a},\) 
since \(A\)  is perfect and the right side is equal to
\({b^{\prime \prime }}^{a}+b^{\prime \prime }\cdot {\left({b}^{b^{\prime }}\right)}^{a}={b^{\prime \prime }}^{a}+{\left({b}^{b^{\prime }}\right)}^{a}.\) 
Since \(A\)  is perfect these two elements commute and the corresponding sums are equal. For the dual actions the equalities are checked similarly.
Now we have to check equality (ii).

For the dot action it is obvious
\({b}^{\left({b^{\prime }}^{b^{\prime \prime }}\right)}\cdot a={b}^{b^{\prime }}\cdot a=a,\) 
for any \(a\in A\)  and any elements from \({\mathbb {b}}^{\left({\mathbb {b}^{\prime }}^{\mathbb {b}^{\prime \prime }}\right)}\)  and \({\mathbb {b}}^{\mathbb {b}^{\prime }}\) .
We have to show
\({a}^{\left({b}^{\left({b^{\prime }}^{b^{\prime \prime }}\right)}\right)}={a}^{\left({b}^{b^{\prime }}\right)}.\) 
The both sides are equal to \({a}^{b}\)  by the definition of action operation in \(\mathsf {Pentact}(A)\) .
It is left to show
\({\left({b}^{\left({b^{\prime }}^{b^{\prime \prime }}\right)}\right)}^{a}={\left({b}^{b^{\prime }}\right)}^{a},\) 
for any elements of \({\mathbb {b}}^{\left({\mathbb {b}^{\prime }}^{\mathbb {b}^{\prime \prime }}\right)}\)  and \({\mathbb {b}}^{\mathbb {b}^{\prime }}\) , and for any \(a\in A\) . The left side is equal to
\({\left({b}^{\left(\left({b^{\prime }}^{b^{\prime \prime }}\right)\cdot a\right)}\right)}^{\left({b^{\prime }}^{b^{\prime \prime }}\right)} = {\left({b}^{a}\right)}^{\left({b^{\prime }}^{b^{\prime \prime }}\right)} = {\left({b}^{a}\right)}^{b^{\prime }}.\) 
The right side is equal to
\({\left({b}^{\left(b^{\prime }\cdot a\right)}\right)}^{b^{\prime }}={\left({b}^{a}\right)}^{b^{\prime }},\) 
by Corollary REF  since \(A\)  is perfect. For the dual actions the corresponding equalities are proved similarly, which ends the proof of Theorem REF .

Proposition 3.15 
Let \(A\)  be a finitely generated abelian group with trivial action on itself. Then there exists a quotient group \(C={A}/{\sim }\)  in \(\mathbf {rGr}^{\bullet }\)  with \(\mathsf {wStab}(C)=0\) .

As we have noted in Section , \(A\)  is an object of \(\mathbf {rGr}^{\bullet }\) . Denote \(\mathsf {wStab}(A)=W_1\)  and \(A_1={A}/{W_1}\) . We have a natural epimorphism \(\tau _1\colon A\rightarrow A_1\) . Suppose \(\mathsf {wStab}(A_1)\ne 0\) . Let \(W_2\)  be a subgroup of \(A\)  generated by the set \(\left\lbrace {\tau }_{1}^{-1}\left(\mathsf {wStab}(A_1)\right)\cup W_1\right\rbrace \) . We have \(W_1\subset W_2\) . Denote \(A_2=A/W_2\)  and let \(\tau _2\colon A\rightarrow A_2\)  be a natural epimorphism. If \(\mathsf {wStab}(A_2)\ne 0\) , let \(W_3\)  be a subgroup of \(A\)  generated by the set \(\left\lbrace {\tau }_{2}^{-1}\left(\mathsf {wStab}(A_2)\right)\cup W_2\right\rbrace \) . We will have \(W_2\subset W_3\)  and if we continue this procedure we will have an increasing sequence
\(W_1\subset W_2\subset W_3\subset \cdots \) 
of subgroups of \(A\) . Since \(A\)  is Noetherian (see Section ), the sequence is finite from which it follows that there exists a number \(n\)  for which there is no \(W_{n+1}\)  in the sequence with \(W_n\subset W_{n+1}\) , which implies that \(\mathsf {wStab}(A_n)=0\) , where \(A_n=A/W_n\) .
Example 3.16 Every finitely generated abelian group gives rise to an object of \(\mathbf {rGr}^{\bullet }\) , which is perfect with zero weak stabilizer. It suffices to apply Proposition REF  and the fact that every abelian group with trivial action on itself is perfect (Example REF ).

Remark 3.17 The idea of constructing objects in \(\mathbf {rGr}^{\bullet }\)  with zero weak stabilizer leads us to the definition and study of Noetherian objects in this category. Note that Noetherian groups are already defined and studied nowadays. One can try to prove like in Proposition REF  that for any Noetherian object in \(\mathbf {rGr}^{\bullet }\)  there exists its quotient object with zero weak stabilizer. It is useful to note that the quotient of a perfect object is also perfect in \(\mathbf {rGr}^{\bullet }\) .

Action of \(\mathsf {Pentact}(A)\)  on \(A\)  and action representability in \(\mathbf {rGr}^{\bullet }\)
Let \(A\in \mathbf {rGr}^{\bullet }\) . An action of \(\mathsf {Pentact}(A)\)  on \(A\)  is defined in a natural way. For any \(\mathbb {b}=\left(b\cdot \text{- },\text{-}\cdot b,{(\text{-})}^{b},{\vphantom{(\text{-})}}^{b}{(\text{-})},{b}^{(\text{-})}\right)\)  we define
\(\mathbb {b}\cdot a=b\cdot a,\) 
which is the value of the function \(b\cdot \text{-}\colon A\rightarrow A\)  on the element \(a\in A\) . Analogously
\({a}^{\mathbb {b}}&={a}^{b}\\{\mathbb {b}}^{a}&={b}^{a}.\) 

By the definition of \(-\mathbb {b}\) , we will have
\((-\mathbb {b})\cdot a &= a\cdot b \\{a}^{-\mathbb {b}} &= {\vphantom{a}}^{b}{a} \\{\left(-\mathbb {b}\right)}^{a} &=-\left({b}^{a}\cdot b\right).\) 
Theorem 4.1 
The action of \(\mathsf {Pentact}(A)\)  on \(A\)  is a derived action in \(\mathbf {rGr}^{\bullet }\)  if \(A\)  is perfect and \(\mathsf {wStab}(A)=0\) , \(A\in \mathbf {rGr}^{\bullet }\) .

We have to check that group action conditions are satisfied, also conditions \((1_{A})-(4_{A})\) , \((1_{B})-(4_{B})\) , \({a}_{1}-{a}_{10}\)  and \({a}^{\mathbb {0}}=a\) , stated in Section , where \(a\in A\) .
Checking of group action conditions,

\(\left(\mathbb {b}+\mathbb {b}^{\prime }\right)\cdot a=\mathbb {b}\cdot \left(\mathbb {b}^{\prime }\cdot a\right)\)  is true by the definition of \(\mathbb {b}+\mathbb {b}^{\prime }\)  and the definition of the action of pentactions on \(A\) .

\(\mathbb {b}\cdot \left(a+a^{\prime }\right)=\mathbb {b}\cdot a+\mathbb {b}\cdot a^{\prime }\) , since it is a property of a pentaction.

\(\mathbb {0}\cdot a=a\)  is true by the definition of the pentaction \(\mathbb {0}\)  and the definition of action of pentactions on \(A\) .

Checking of conditions \((1_{A})-(4_{A})\) ,

This condition is true by the definition of pentaction, condition 2.

This condition is true by the definition of the sum \(\mathbb {b}+\mathbb {b}^{\prime }\)  of pentactions.

We have \({\left(\mathbb {b}\cdot a\right)}^{a^{\prime }}={a}^{a^{\prime }}\)  for any \(a,a^{\prime }\in A\) , with \(a^{\prime }\ne 0\)  by the definition of pentaction, condition 3.

\({\left(\mathbb {b}\cdot a\right)}^{\mathbb {b}^{\prime }}={a}^{\mathbb {b}^{\prime }}\)  is true by the definition of action of \(\mathsf {Pentact}(A)\)  on \(A\)  and Lemma REF , since \(A\)  is perfect.

Checking of conditions \((1_{B})-(4_{B})\) ,

This condition is true by the definition of pentaction, condition 4.

This condition is true by the definition of the sum \(\mathbb {b}+\mathbb {b}^{\prime }\) .

This condition is true by the definition of pentaction, property 5.

This condition is true by the definition of \({\mathbb {b}}^{\mathbb {b}^{\prime }}\) .

Obviously \({a}^{\mathbb {0}}=a\) , for any \(a\in A\) .
Checking of the conditions \({a}_{1}-{a}_{10}\) .

Condition \({a}_{1}\)  is true by Definition REF , property 6.

This condition is true since \(A\)  is perfect.

This condition is true by the definition of \({\mathbb {b}}^{\mathbb {b}^{\prime }}\) .

This condition is true by Definition REF , property 7.

This condition is true by the definition of \({\mathbb {b}}^{\mathbb {b}^{\prime }}\) .

This condition is true by Definition REF , condition 8.

This condition is true by Definition REF , condition 9.

This condition is true by Definition REF , condition 10.

\({\mathbb {b}}^{\left({\mathbb {b}^{\prime }}^{a}\right)}=0\)  is true since \(\mathsf {wStab}(A)=0\) .

\({\mathbb {b}}^{\left({a}^{\mathbb {b}^{\prime }}\right)}={\mathbb {b}}^{a}\)  is true since \(\mathsf {wStab}(A)=0\) .

Therefore all conditions of derived actions in \(\mathbf {rGr}^{\bullet }\)  are satisfied, which proves the theorem.
Theorem 4.2 
Let \(A\in \mathbf {rGr}^{\bullet }\) . If \(A\)  is perfect and \(\mathsf {wStab}(A)=0\) , then \(A\)  is action representable and
\(\mathbb {AR}(A)=\mathsf {Pentact}(A).\) 

Let \(B\in \mathbf {rGr}^{\bullet }\)  which has a derived action on \(A\) . Define a function
\(\varphi \colon B\rightarrow \mathsf {Pentact}(A)\) 
in the following way. For any \(b\in B\)
\(\varphi (b)=\left(b\cdot \text{- }, \text{-}\cdot b, {(\text{-})}^{b}, {\vphantom{(\text{-})}}^{b}{(\text{-})}, {b}^{(\text{-})}\right),\) 
where \(b\cdot a\)  is an action of \(b\)  on \(a\) ; \(a\cdot b\)  is an action of \(-b\)  on \(a\) , i.e. \(a\cdot b=(-b)\cdot a\) ; \({a}^{b}\)  is an action of \(b\)  on \(a\) ; \({\vphantom{a}}^{b}{a}={a}^{-b}\) , where on the right side is the action of \(-b\)  on \(a\) ; \({b}^{a}\)  is an action of \(b\)  on \(a\) , for any \(a\in A\) . Easy checking shows that \(\varphi (b)\)  is a pentaction of \(A\) ; \(\varphi \)  is a homomorphism of \(\mathbf {rGr}^{\bullet }\) , it follows from the definition of the operations in \(\mathsf {Pentact}(A)\)  and the construction of \(\varphi \) . We have
\(\begin{alignedat}{2}b\cdot a &= \varphi (b)\cdot a \\{a}^{b} &= {a}^{\varphi (b)} \\{b}^{a} &= {\varphi (b)}^{a}.\end{alignedat}\) 
\(\varphi \)  is unique function with properties (REF ), which follows from the definition of pentactions and conditions (REF ).

Acknowledgements
It was the second author's idea to define and describe actions in the category of groups with action. It was stimulating and then interesting to search such a subcategory in this category, where it would be possible to investigate action representability of certain objects.
